# The upper 3/4th portion of a vertical pole subtends an angle $\dpi{100} tan^{-1}(3/5)$   at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is Option 1) 40 m Option 2) 60 m Option 3) 80 m Option 4) 20 m

As we learnt in

Height and Distances -

The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.

-

$\Theta =\tan ^{-1}\frac{3}{5}$

Now $\tan\Theta =\tan (\angle ACB-\angle DCB)$

$= \frac{\tan(\angle ACB-\angle DCB)}{1+\tan (\angle ACB)\tan \angle DCB}\, \, \, \, \,......(i)$

In $\Delta ACB, \tan \angle ACB=\frac{h}{BC}$                -----(ii)

Similarly, $\Delta DCB, \tan \angle DCB=\frac{h}{4BC}$

using (i) & (ii)

$\tan \Theta =\frac{3}{5} = \frac{\frac{h}{BC}-\frac{h}{4BC}}{1+\frac{h^{2}}{4(BC)^{2}}}$$=\frac{BC[3h]}{4BC^{2}+h^{2}}$

$\Rightarrow \frac{3}{5}=\frac{120h}{6400+h^{2}}$

$\Rightarrow 3h^{2}-600h+19200=0$

solving, $h=40\ or\ h=160$

Option 1)

40 m

This is correct option

Option 2)

60 m

This is incorrect option

Option 3)

80 m

This is incorrect option

Option 4)

20 m

This is incorrect option

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
##### Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-